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I've been thinking about this for a while, and can't seem to find any way to do it despite the statement itself seeming obvious. The problem is:

Let $f:[0,1] \to \mathbb{R}^n$ be a continuous map, not necessarily injective, such that $f(0) \not = f(1)$. Let $Y$ denote the image of $f$ as a compact subset of $\mathbb{R}^n$. Then there exists an injective map $g:[0,1] \to Y$ such that $g(0) = f(0)$ and $g(1) = f(1)$.

Obviously, the problem is trivial in many cases, but gets tough when you consider "wild" curves. Below I'll discuss how I've tried to solve it, but feel free to ignore it if you know the answer.

My first idea was to start Zorn's lemma argument using a sequence of maps $f_0,f_1,f_2,f_3,\dots$ with $f_0 = f$ and $f_0([0,1]) \supseteq f_1([0,1]) \supseteq f_2([0,1]) \supseteq \cdots$ and trying to find a bound for the chain, but it seems this fails because given compact path-connected subsets $X_0,X_1,\dots$ where $X_0 \supseteq X_1 \supseteq \cdots$, the intersection $\bigcap_{k=1}^\infty X_k$ is not necessarily path connected. For example, if $S \subset \mathbb{R}^2$ is a segment of the (closed) topologist's sine curve, then the family $S \cup \bar{B}_{1/n}(0)$ provide a counterexample.

The other approach seemed to be trying to define a sequence of functions $f_0,f_1,\dots$ where each $f_n$ is "closer" to being injective than the one before it, and such that the sequence converges in some meaningful way to an injective function, or at least to a function that allows us to use a simpler method to finish the problem. However, I don't really know enough analysis to follow through with this approach, so I'm asking for help here. Any solution would be great, and solutions a method similar to what I've mentioned above would be doubly appreciated.

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If Y is an auto intersecting loop, if we take out the intersection point, we divide it in three connected components. Then there is no injective continuous function from [0,1] to Y, as the image can have a maximum of 2 components. –  juanrapha Jun 8 '12 at 6:06
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@JuanSimões: I don't know whether I understand, but it seems to me like you are talking about the nonexistence of a bijective function from $[0,1]$ to $Y$. This is not what the question asks us to do. All that Carl wants to construct is an injective path in $Y$ connecting $f(0)$ to $f(1)$. –  Sam Jun 8 '12 at 6:45
    
Can the length of $f$ be infinite? If not, then such $g$ is found with very simple reparameterization argument. –  Thomas E. Jun 8 '12 at 10:36
    
@Thomas: $f$ could be a space-filling curve. –  Brian M. Scott Jun 8 '12 at 10:44

1 Answer 1

up vote 3 down vote accepted

It suffices to show that $Y$ is arcwise connected, since a homeomorphism $g:[0,1]\to Y$ such that $g(0)=f(0)$ and $g(1)=f(1)$ is certainly injective.

The Hahn-Mazurkiewicz theorem says that a Hausdorff space is the continuous image of the closed unit interval iff it is a compact, connected, locally connected metric space; such spaces are sometimes called Peano spaces. Clearly $Y$ is a Peano space. The result is therefore immediate from the theorem that every Peano space is arcwise connected. This is Theorem 31.2 in Stephen Willard, General Topology; the proof is non-trivial but not hard to follow.

You may be able to see most of it at Google Books, if you you’re allowed to read the page 220 result here. The missing lines from page 219 (including the end of the sentence from p. 220) are:

Suppose $a$ and $b$ are points in a Peano space $X$. Using Theorem 26.15, there is a simple chain $U_{11},\dots,U_{1n}$ of open connected sets of diameter $<1$ from $a$ to $b$.

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Very nice solution. –  copper.hat Jun 8 '12 at 16:18

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